sqing wrote: Let $ a_1, a_2,..., a_n $ be non-negetive numbers , prove that \[\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+...\] \[+\frac{ a_1 a_2... a_{n-1}}{(1+ a_1)(1+ a_2)...(1+ a_n)} \le 1\] $ 1-\frac{1}{1+a_1}=\frac{a_1}{1+a_1};1-\frac{1}{1+a_1}-\frac{a_1}{(1+a_1)(1+a_2)}=\frac{a_1a_2}{(1+a_1)(1+a_2)}$ ... $1-(\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+... +\frac{ a_1 a_2... a_{n-1}}{(1+ a_1)(1+ a_2)...(1+ a_n)})=\frac{a_1a_2...a_n}{(1+a_1)(1+a_2)...(1+a_n)}\ge0 $ The equality holds iff $a_1=0$ or $a_2=0$ or...or $a_n=0$

sqing wrote: Let $ a_1, a_2,..., a_n $ be non-negetive numbers , prove that \[\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+...\] \[+\frac{ a_1 a_2... a_{n-1}}{(1+ a_1)(1+ a_2)...(1+ a_n)} \le 1\] \[\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+...\] \[+\frac{ a_1 a_2... a_{n-1}}{(1+ a_1)(1+ a_2)...(1+ a_n)} \] \[\le \frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+...\] \[+\frac{ a_1 a_2... a_{n-1}}{(1+ a_1)(1+ a_2)...(1+ a_{n-1})}=1 \] Equality holds when $a_n=0$

Proof: \[\frac{1}{1+{{a}_{1}}}+\frac{{{a}_{1}}}{(1+{{a}_{1}})(1+{{a}_{2}})}+\frac{{{a}_{1}}{{a}_{2}}}{(1+{{a}_{1}})(1+{{a}_{2}})(1+{{a}_{3}})}+...+\frac{{{a}_{1}}{{a}_{2}}...{{a}_{n-1}}}{(1+{{a}_{1}})(1+{{a}_{2}})...(1+{{a}_{n}})}=\] \[\frac{1}{1+{{a}_{1}}}+\sum\limits_{k=2}^{n}{\frac{{{a}_{1}}{{a}_{2}}...{{a}_{k-1}}}{(1+{{a}_{1}})(1+{{a}_{2}})...(1+{{a}_{k}})}}=\frac{1}{1+{{a}_{1}}}+\sum\limits_{k=2}^{n}{\frac{{{a}_{1}}{{a}_{2}}...{{a}_{k-1}}\left( 1+{{a}_{k}}-{{a}_{k}} \right)}{(1+{{a}_{1}})(1+{{a}_{2}})...(1+{{a}_{k}})}}=\] \[\frac{1}{1+{{a}_{1}}}+\sum\limits_{k=2}^{n}{\left( \frac{{{a}_{1}}{{a}_{2}}...{{a}_{k-1}}}{(1+{{a}_{1}})(1+{{a}_{2}})...(1+{{a}_{k-1}})}-\frac{{{a}_{1}}{{a}_{2}}...{{a}_{k}}}{(1+{{a}_{1}})(1+{{a}_{2}})...(1+{{a}_{k}})} \right)}=\] \[\frac{1}{1+{{a}_{1}}}+\frac{{{a}_{1}}}{1+{{a}_{1}}}-\frac{{{a}_{1}}{{a}_{2}}...{{a}_{n}}}{(1+{{a}_{1}})(1+{{a}_{2}})...(1+{{a}_{n}})}\le 1.\]

Generalization Let $a_1, a_2,..., a_n$ be non-negative numbers and $f:\left[ 0,\infty \right)\to \left[ 0,\infty \right),g:\left[ 0,\infty \right)\to \left( 0,\infty \right)$, prove that \[\frac{g\left( {{a}_{1}} \right)-f\left( {{a}_{1}} \right)}{g\left( {{a}_{1}} \right)}+\frac{f\left( {{a}_{1}} \right)}{g\left( {{a}_{1}} \right)g\left( {{a}_{2}} \right)}\left( g\left( {{a}_{2}} \right)-f\left( {{a}_{2}} \right) \right)+\frac{f\left( {{a}_{1}} \right)f\left( {{a}_{2}} \right)}{g\left( {{a}_{1}} \right)g\left( {{a}_{2}} \right)g\left( {{a}_{3}} \right)}\left( g\left( {{a}_{3}} \right)-f\left( {{a}_{3}} \right) \right)+...\] \[+\frac{f\left( {{a}_{1}} \right)f\left( {{a}_{2}} \right)...f\left( {{a}_{n-1}} \right)}{g\left( {{a}_{1}} \right)g\left( {{a}_{2}} \right)...g\left( {{a}_{n}} \right)}\left( g\left( {{a}_{n}} \right)-f\left( {{a}_{n}} \right) \right)\le 1.\]

(Short list IMO,1998) Let $ a_1, a_2,..., a_n $ be positive numbers with $ a_1+a_2+...+a_n<1 $, prove that \[\frac{ a_1 a_2... a_n}{(1- a_1)(1- a_2)...(1- a_n)} \le \frac{a_1+a_2+...+a_n}{n^{n+1}(1-(a_1+a_2+...+a_n))}.\]

sqing wrote: (Short list IMO,1998) Let $ a_1, a_2,..., a_n $ be positive numbers with $ a_1+a_2+...+a_n<1 $, prove that \[\frac{ a_1 a_2... a_n}{(1- a_1)(1- a_2)...(1- a_n)} \le \frac{a_1+a_2+...+a_n}{n^{n+1}(1-(a_1+a_2+...+a_n))}.\] $a_{n+1}=1-\sum_{i=1}^{n}a_i;a_{n+1}>0$ The desired inequality becomes $\frac{ a_1 a_2... a_na_{n+1}}{(1- a_1)(1- a_2)...(1- a_n)(1- a_{n+1})} \le \frac{1}{n^{n+1}}$ $1-a_i=a_1+...+a_{i-1}+a_{i+1}+...+a_{n+1}\ge n\sqrt[n]{a_1...a_{i-1}a_{i+1}...a_{n+1}}$ Done.

I think the problem is related to Probability. See here. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=498&t=493730

We claim \[ \frac {1}{a_1+1}+\frac {a_1}{(1+a_1)(1+a_2)}+.....+\frac {a_1a_2.....a_{n-1}}{(1+a_1)(1+a_2).....(1+a_n)}\] \[=1-\frac{a_1a_2.....a_n}{(1+a_1)(1+a_2).....(1+a_n)} \] We'll prove this using mathematical induction. For $n=1,(1)$ is true since $\frac {1}{1+a_1}=1-\frac {a_1}{1+a_1}$ Let $(1)$ be true for some $m \in \mathbb {N}$ So \[\frac {1}{1+a_1}+\frac {a_1}{(a_1+1)(a_2+1)}+.....+\frac {a_1a_2.....a_{m-1}}{(1+a_1)(1+a_2).....(1+a_m)}\] \[=1-\frac {a_1a_2.....a_m}{(1+a_1)(1+a_2).....(1+a_m)}\] Now for $m+1$, \[\frac {1}{1+a_1}+\frac {a_1}{(a_1+1)(a_2+1)}+.....+\frac {a_1a_2.....a_{m-1}}{(1+a_1)(1+a_2).....(1+a_m)}+\frac {a_1a_2.....a_m}{(1+a_1)(1+a_2).....(1+a_{m+1})}\] \[=1-\frac {a_1a_2.....a_m}{(1+a_1)(1+a_2).....(1+a_m)}+\frac {a_1a_2.....a_m}{(1+a_1)(1+a_2).....(1+a_{m+1})}\] \[=1-\frac {a_1a_2.....a_{m+1}}{(1+a_1)(1+a_2).....(1+a_{m+1})}\] So by mathematical induction $(1)$ holds $\forall n \in \mathbb {N}$. From $(1)$ we get $\frac {1}{a_1+1}+\frac {a_1}{(1+a_1)(1+a_2)}+.....+\frac {a_1a_2.....a_{n-1}}{(1+a_1)(1+a_2).....(1+a_n)} \le 1$ because $\frac{a_1a_2.....a_n}{(1+a_1)(1+a_2).....(1+a_n)} \ge 0$ as $a_1,a_2,.....,a_n$ are non-negative with equality iff at least one of $a_1,a_2,.....,a_n$ is equal to $0$

Solution: Let $b_1=(1+a_2)...(1+a_n), b_n=a_1a_2...a_{n-1}, b_{n+1}=a_1a_2...a_{n}$ and $b_i=a_1a_2...a_{i-1}(1+a_{i+1})...(1+a_n)$ for $2\leq i\leq n-1$.Furthermore let $B=(1+a_1)(1+a_2)...(1+a_n)$.The inequality can be re-written as \[A=\sum _{k=1}^n b_k\leq B\]Now notice that \[A+b_{n+1}=\sum _{k=1}^{n-1} b_k+(1+a_n)b_n=\sum _{k=1}^{n-2} b_k+a_1a_2...a_{n-2}(1+a_n)+(1+a_n)b_{n}=\sum _{k=1}^{n-2} b_k+(1+a_{n-1})b_{n-1}\] By continuing this argument at the last we get $A+b_{n+1}=(1+a_1)b_1=B$ and so we are done.The equality occurs if $b_{n+1}=0$ i.e. some of the $a_i$s is equal to zero.

Solution by Kuronyanko, Japan. Let $Q_n=(1+a_1)(1+a_2)\ \cdots (1+a_n)$ $P_n=a_1a_2\cdots a_n$ $Q_0=1,\ P_0=1$. We can rephrase the given inequality as \[\frac{P_0}{Q_1}+\frac{P_1}{Q_2}+\cdots +\frac{P_{n-1}}{Q_n}\leq 1\ \cdots [*]\] Proof : $1-\frac{P_0}{Q_1}=\frac{a_1P_0}{Q_1}=\frac{P_1}{Q_1}$, $\frac{P_1}{Q_1}-\frac{P_1}{Q_2}=\frac{a_2P_1}{Q_2}=\frac{P_2}{Q_2}.$ Generally, $\frac{P_i}{Q_i}-\frac{P_i}{Q_{i+1}}=\frac{a_{i+1}P_i}{Q_{i+1}}=\frac{P_{i+1}}{Q_{i+1}}.$ Summing up, from $i=1$ through $n-1$, yielding $1-\sum_{i=0}^{n-1} \frac{P_i}{Q_{i+1}}=\frac{P_n}{Q_n}\geq 0.$ Equality holds when $a_i=0$ for some $i$.

You could take a a1/(a1+1) out of the last n-1 terms and use induction.

Let $a_1,a_2,\dots,a_n$ be positive real numbers whose product is $1$. Show that \[ \frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+...+\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} \ge \frac{2^n-1}{2^n} .\] 2014 CMO

$\frac{a_1}{1+a_1}+.......+\frac{a-n}{(1+a_1)....(1+a_n)}\$ $= 1-(1/1+a_1)+......+ (1/(1+a_1)....(1+a_n-1)- (1/(1+a_1)....(1+a_n))$ hence,$\[ \frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+...+\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}$ = $1- 1/(1+a_1)....(1+a_n)$ now to prove that $\[ \frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+...+\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}\ge\frac{2^n-1}{2^n}. \]$ it reduces to prove $1- 1/(1+a_1)....(1+a_n)ge\frac{2^n-1}{2^n}. $ now this reduces to proving $1/(1+a_1)...(1+a_n) <= (1/2^n)$ or $(1+a_1)(1+a_2)....(1+a_n) >= 2^n$ which easily follows by AM $>=$ GM since $1+a1 >= 2 a_1^1/2$ ..... $1+a_n >= 2xa_n^1/2$ hence proved!

Note that $a_n$ only appears in the denominator, so we can assume it is 0 (to get a maximum). $ \frac{1}{1+a_1}+\frac{ a_1}{(1+a_1)(1+a_2)}+\frac{ a_1 a_2}{(1+a_1)(1+a_2)(1+a_3)}+ $ $ \cdots+\frac{ a_1 a_2\cdots a_{n-1}}{(1+a_1)(1+a_2)\cdots (1+a_n)}$ $=\frac{1}{1+a_1}+\frac{ a_1}{(1+a_1)(1+a_2)}+\frac{ a_1 a_2}{(1+a_1)(1+a_2)(1+a_3)}+ $ $ \cdots+\frac{ a_1 a_2\cdots a_{n-3}}{(1+a_1)(1+a_2)\cdots (1+a_{n-2})}+\frac{ a_1 a_2\cdots a_{n-2}}{(1+a_1)(1+a_2)\cdots (1+a_{n-1})}+\frac{ a_1 a_2\cdots a_{n-1}}{(1+a_1)(1+a_2)\cdots (1+a_{n-1})}$ $=\frac{1}{1+a_1}+\frac{ a_1}{(1+a_1)(1+a_2)}+\frac{ a_1 a_2}{(1+a_1)(1+a_2)(1+a_3)}+ $ $ \cdots+\frac{ a_1 a_2\cdots a_{n-3}}{(1+a_1)(1+a_2)\cdots (1+a_{n-2})}+\frac{ a_1 a_2\cdots a_{n-2}}{(1+a_1)(1+a_2)\cdots (1+a_{n-2})}$ $= \cdots=\frac{1}{1+a_1}+\frac{a_1}{1+a_1}=1$, as required.

Let $a_{i}=\tan^2(x_{i}), x_{i} \in (0, \pi /2)$, using $\frac{1}{1+\tan^2(x)}=\cos^2(x)$, inequality transform to $\sum_{i=1}^{n}\frac{a_{1}a_{2}\ldots s_{i-1}}{(1+a_{1})(1+a_{2})\ldots (1+a_i{})}$ = $\sum_{i=1}^{n}\sin^2(x_{1})\sin^2(x_{2})\ldots \sin^2(x_{i-1})\cos^2(x_{i})$ = $\sum_{i=1}^{n}(\sin^2(x_{1})\sin^2(x_{2})\ldots \sin^2(x_{i-1}) - \sin^2(x_{1})\sin^2(x_{2})\ldots \sin^2(x_{i-1})\sin^2(x_{i}))$ = $1 - \sin^2(x_{1})\sin^2(x_{2})\ldots \sin^2(x_{n}) \leq 1$, which is obvius.

Think this solution is slightly different from others, may be wrong?

This problem is absolutely hilarious.

We proceed to prove this by induction. Base Case: $n=1$ $$\frac{1}{1+a_1}\leq 1\iff 1\leq 1+a_1\iff a_1\geq 0$$which is true by condition. Assume that for some positive integer $k$, we have that $$\frac{1}{1+a_1}+\cdots +\frac{a_1\cdots a_{k-1}}{(1+a_1)\cdots (1+a_k)}\leq 1$$note that this is the same as $$\frac{1}{1+a_1}(1+\frac{a_1}{1+a_2}(\cdots (1+\frac{a_{n-2}}{1+a_{n-1}}(1+\frac{a_{n-1}}{1+a_n}))\cdots))\leq 1$$Then notice that $$\frac{1}{1+a_1}+\cdots +\frac{a_1\cdots a_k}{(1+a_1)\cdots (1+a_{k+1})} $$$$= \frac{1}{1+a_1}(1+\frac{a_1}{1+a_2}(\cdots (1+\frac{a_{k-1}}{1+a_k}(1+\frac{a_k}{1+a_{k+1}}))\cdots)) \leq 1$$$$\iff 1+\frac{a_1}{1+a_2}(\cdots (1+\frac{a_{k-1}}{1+a_k}(1+\frac{a_k}{1+a_{k+1}}))\cdots)\leq 1+a_1$$$$\iff \frac{a_1}{1+a_2}(\cdots (1+\frac{a_{k-1}}{1+a_k}(1+\frac{a_k}{1+a_{k+1}}))\cdots) \leq a_1$$Then we have two cases, if $a_1 = 0$, we are obviously done with $0\leq 0$, if it's not $0$, we continue, $$\iff \frac{1}{1+a_2}(\cdots (1+\frac{a_{n-1}}{1+a_n}(1+\frac{a_n}{1+a_{n+1}}))\cdots)\leq 1$$However here, by setting $$a_1\to a_2\;\;a_2\to a_3\;\cdots\;a_{k}\to a_{k+1}$$in the inductive hypothesis, we have the desired result, and we complete our induction.

The main difficulty of this problem is observing that no inequality is involved. We will prove by induction that the sum is equal to $1-\frac{\prod_{i=1}^{n}a_i}{\prod_{i=1}^{n}(a_i+1)}$, which also establishes the equality case as when $\prod\limits_{i=1}^{n}a_i=0$. The base case, $n=1$, is trivial. For the inductive step, notice that $\frac{\prod_{i=1}^{n}a_i}{\prod_{i=1}^{n}(a_i+1)}=\frac{(\prod_{i=1}^{n}a_i)(a_{n+1}+1)}{\prod_{i=1}^{n+1}(a_i+1)}$ and $\frac{(\prod_{i=1}^{n}a_i)(a_{n+1}+1)}{\prod_{i=1}^{n+1}(a_i+1)}-\frac{(\prod_{i=1}^{n}a_i)}{\prod_{i=1}^{n+1}(a_i+1)}=\frac{(\prod_{i=1}^{n}a_i)(a_{n+1}+1-1)}{\prod_{i=1}^{n+1}(a_i+1)}=\frac{(\prod_{i=1}^{n+1}a_i)}{\prod_{i=1}^{n+1}(a_i+1)}$.

Induct on $n$. The base case $n=1$ is trivial; for the inductive step, note that \begin{align*} \frac 1{a_1+1} + \frac{a_1}{(1+a_1)(1+a_2)}+\cdots+\frac{a_1a_2\cdots a_{n-1}}{(1+a_1)(1+a_2) \cdots (1+a_n)} &\leq 1 \\ \iff 1+\frac{a_1}{1+a_2}+\cdots+\frac{a_1a_2\cdots a_{n-1}}{(1+a_2)(1+a_3)\cdots (1+a_n)} &\leq 1+a_1 \\ \iff \frac 1{1+a_2} + \frac{a_2}{(1+a_2)(1+a_3)}+\cdots+\frac{a_2a_3\cdots a_{n-1}}{(1+a_2)(1+a_3)\cdots (1+a_n)} &\leq 1, \end{align*}which is just the case for $n-1$ variables.

Let prove this inequality by induction. Let $\P(x)=\frac{a_1a_2\cdots a_{x-1}}{(1+a_1)(1+a_2)\cdots(1+a_x)}$ $\textcolor{red}{Claim:}$ For any positive integer $n$ \[ \sum_{i=1}^n \left( \P(i) \right)=1-\frac{a_1a_2\cdots a_{n+1}}{(1+a_1)(1+a_2)\cdots(1+a_{n+2})}\]For $n=1$ it is trivial: $\frac {1}{1+a_1}=1-\frac {a_1}{1+a_1}$ Assume that this is true for some $m$ $\implies$ \[ \sum_{i=1}^m \left( \P(i) \right)=1-\frac{a_1a_2\cdots a_{m+1}}{(1+a_1)(1+a_2)\cdots(1+a_{m+2})}\]Now for $m+1$, \[\frac {1}{1+a_1}+.....+\frac {a_1a_2.....a_{m+1}}{(1+a_1)(1+a_2).....(1+a_{m+2)}}+\frac {a_1a_2.....a_{m+2}}{(1+a_1)(1+a_2).....(1+a_{m+3})}\]\[=1-\frac {a_1a_2.....a_{m+2}}{(1+a_1)(1+a_2).....(1+a_{m+2})}+\frac {a_1a_2.....{a_{m+2}}}{(1+a_1)(1+a_2).....(1+a_{m+3})}\]\[=1-\frac {a_1a_2.....a_{m+3}}{(1+a_1)(1+a_2).....(1+a_{m+3})}\]Which means that \[ \sum_{i=1}^{n+1} \left( \P(i) \right)=1-\frac{a_1a_2\cdots a_{n+2}}{(1+a_1)(1+a_2)\cdots(1+a_{n+3})}\]is also true. So our claim is true for all positive integer $n$. Equality holds if $a_n = 0$.

For $1 \le i \le n$, let $p_i = \tfrac{1}{1 + a_i}$, so $0 < p_i \le 1$. We now wish to prove that $$p_1 + (1 - p_1)p_2 + (1 - p_1)(1 - p_2)p_3 + \cdots + (1 - p_1)(1 - p_2) \cdots (1 - p_{n - 1})p_n \le 1.$$Now, consider $n$ coins $C_1, C_2, \ldots, C_n$ such that the probability of $C_i$ flipping heads is $p_i$. Then the probability of eventually getting heads if we first flip $C_1$, then $C_2$, and so on and stop after we flips heads is the sum in question; but obviously this is less than or equal to $1$. Equality holds when the probability of any of the coins flipping heads is $1$, i.e. at least one of the $a_i$ equals $0$.

Noice. Consider adding all fractions on the LHS, and the numerator of the sum. Note that for $n = 2$ it is of the form, \[ (1+a_1) + a_1(1+a_1)(1+a_2) = (1 + a_1)(1+a_2) - a_1a_2 \]Now assume for $n = k$ we have, \begin{align*} (1+a_2)\dots(1+a_{k}) + a_1(1+a_2)\dots&(1+a_{k}) + \dots + a_1a_2\dots a_{k}\\ &= (1+a_1)(1+a_2)\dots(1+a_{k}) - a_1a_2\dots a_{k} \end{align*}Then for $n = k+1$ we wish to have, \begin{align*} (1+a_2)\dots(1+a_{k+1}) + a_1(1+a_2)\dots&(1+a_{k+1}) + \dots + a_1a_2\dots a_{k+1}\\ &= (1+a_1)(1+a_2)\dots(1+a_{k+1}) - a_1a_2\dots a_{k+1} \end{align*}The LHS can be written as, \begin{align*} (1+a_{k+1}) \cdot [(1+a_1)(1+a_2)\dots &(1+a_k) - a_1a_2\dots a_k]\\ &= (1+a_1)(1+a_2)\dots (1+a_{k+1}) - a_1a_2\dots a_{k+1} \end{align*}which completes our induction as desired. Now from our induction we have, \[ \frac{1}{1+a_1} + \frac{a_1}{(1+a_1)(1+a_2)} + \cdots + \frac{a_1a_2\cdots a_{n-1}}{(1+a_1)(1+a_2)\cdots(1+a_n)} = 1 - \frac{a_1a_2\dots a_n}{(1+a_1)(1+a_2)\dots (1+a_n)}.\] Thus the fraction is always less than 1 with equality holding if and only if some $a_i = 0$.

Counting + Algebra = ???? Define $x_i = \frac{1}{1+a_i},$ so that $0 < x_i \le 1.$ The problem is then equivalent to $$x_1 + (1-x_1) x_2 + (1-x_1)(1-x_2) x_3 + \dots \le 1.$$Now, create $n$ coins, where the $i$th coin has a probability of $x_i$ to come up heads. Consider the sequence of actions of flipping all of the $n$ coins in order until we get a coin that flips heads. The above expression is then the probability that we see one of the coins come up heads, which is clearly less than or equal to $1.$ Equality holds if and only if one of the coins always comes up heads, i.e. at least one of the $x_i$ is equal to $1.$ This is equivalent to at least one of the $a_i$ being equal to $0,$ and we are done.

It is sufficient to show that \[1-\left(\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\cdots+\frac{ a_1 a_2\cdots a_{n-1}}{(1+ a_1)(1+ a_2)\cdots (1+ a_n)}\right)=\frac{a_1a_2\cdots a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}\]But this follows from induction on $n$ as \[\frac{a_1a_2\cdots a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}-\frac{a_1a_2\cdots a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)(1+a_{n+1})}=\frac{a_1a_2\cdots a_na_{n+1}}{(1+a_1)(1+a_2)\cdots(1+a_n)(1+a_{n+1})}\]

Clearly, $\tfrac{1}{1+a_n} \leq 1$. Then: \begin{align*} \dfrac{a_{n - 1}}{1 + a_n} \leq a_{n-1} \quad \longrightarrow \quad 1 + \dfrac{a_{n - 1}}{1 + a_n} \leq 1 + a_{n-1} \quad \longrightarrow \quad \dfrac{1}{1+a_{n-1}} + \dfrac{a_{n-1}}{\left(1+a_n\right)\left(1+a_{n-1}\right)} \leq 1. \end{align*}Repeating the process, we are guaranteed to obtain the final result.

If $S_i$ is the $i$th elementary symmetric polynomial of the roots, it is possible to show by induction that $$\text{LHS} = \frac{S_0+S_1+\cdots+S_{n-1}}{\prod_{i=1}^n (1+a_i)}.$$Therefore, we have that $$\text{LHS} = \frac{\prod_{i=1}^n (1+a_i) - S_n}{\prod_{i=1}^n (1+a_i)} = 1 - \prod_{i=1}^n \frac{a_i}{1+a_i} \leq 1,$$so the inequality is proven. From here, it is clear that equality holds if and only if at least one of the $a_i=0.$ QED

why can't more ineqs be like this \[\frac{1}{1+a_1} + \frac{a_1}{(1+a_1)(1+a_2)} + \cdots + \frac{a_1a_2\cdots a_{n-1}}{(1+a_1)(1+a_2)\cdots(1+a_n)} \]\[= 1-\frac{a_1}{1+a_1} + \frac{a_1}{1+a_1}-\frac{a_1a_2}{(1+a_1)(1+a_2)} + \cdots + \frac{a_1a_2\cdots a_{n-1}}{(1+a_1)(1+a_2)\cdots(1+a_{n-1})}-\frac{a_1a_2\cdots a_{n-1}a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}\]\[= 1-\frac{a_1a_2\cdots a_{n-1}a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}.\]This is clearly $\le 1$ with equality case when any of the variables is $0$.